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小木虫论坛-学术科研互动平台 » 计算模拟区 » 计算模拟 » 求解高次复杂函数
北京石油化工学院2026年研究生招生接收调剂公告
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denmarkrico

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[求助] 求解高次复杂函数 已有3人参与

求解下面这个高次方程时为什么总是得到虚数解，我所用的是MATLAB2012b版

syms h c l theta11 theta21 theta31 x
h=14.5;
l=55.5;
c=19.73;
theta11=0.01/37;
theta21=0.035/37;
theta31=0.065/37;
F=((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +3^(1/2)*c*sin(theta0) - c*cos(theta0) +l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + (((433*c)/250 + (433*h)/250 + 3*l + 2*l*sin(theta11) + 2*l*sin((2*pi)/3 + theta21) - (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) - (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)) - ((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + ((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))^2*(2*c*h - 2*l^2*(sin(theta11) - 1) + 2*c^2 - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h)) + ((((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + ((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*c + 2*h + 2*l*cos(theta11) + (2*c*(x^2 - 1))/(x^2 + 1)) - ((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*l*sin(theta11) - 2*l + (4*c*x)/(x^2 + 1)))*((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) - l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))；
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心 那个凉啊
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denmarkrico

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2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

theta0 是要求的量
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心 那个凉啊
8楼2013-12-14 16:09:58
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feixiaolin

无虫

【答案】应助回帖

感谢参与，应助指数 +1
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？
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denmarkrico
2楼2013-12-13 22:53:29
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jerkwin

新虫
你这方程, 能解出来我觉得很奇怪
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3楼2013-12-13 23:10:14
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paopaoai11

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你这什么方程啊？

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泡泡
4楼2013-12-14 01:03:00
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